Question
What volume of oxygen gas at STP would be needed for the complete combustion of 39.0 grams of octane ?
Answers
22.4 L/mole ... gas at stp
octane is 114 g/mole
2 C8H18(g) + 25 O2(g) ---> 16 CO2 (g) + 18 H2O(l)
(39.0 / 114) * 12.5 * 22.4 L
octane is 114 g/mole
2 C8H18(g) + 25 O2(g) ---> 16 CO2 (g) + 18 H2O(l)
(39.0 / 114) * 12.5 * 22.4 L
where did you get the 12.5
there are 25 O2 so there are 12.5 O
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